An nfl linebacker can go from 0 m/s to 2.5 m/s in 2.5s. What is his acceleration?
1 answer:
You might be interested in
Answer:
Range,
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}
Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;
Where;
d is the separation or distance between the two parallel plates;
Therefore, the necessary separation between the two parallel plates is 0.104 mm
Answer:
Approximate height of the building is 23213 meters.
Explanation:
Let the height of the building be represented by h.
0.02 radians = 0.02 ×
= 0.02 x (180/)
0.02 radians = 1.146°
10.5 km = 10500 m
Applying the trigonometric function, we have;
Tan θ =
So that,
Tan 1.146° =
⇒ h = Tan 1.146° x 10500
= 2.21074 x 10500
= 23212.77
h = 23213 m
The approximate height of the building is 23213 m.